Let the two roots be $\alpha$ and $\beta$ as follows:

$\alpha = |3-4i|=5$

$\beta = \omega^2+\omega$ and since $\omega^2+\omega +1 =0 \quad \Rightarrow \omega^2+\omega = \beta = -1$

By Vieta Formulas the equation must be:

$x^2-(\alpha+\beta)x+(\alpha \beta) = 0\quad \Rightarrow \boxed {x^2-4x-5=0}$

i did forget what was meant by |3-4i| but omega^2+omega was -1 and thus i managed to find the roots of all equations!

What i was about to do?

Report the question as wrong as imaginary roots occur in pairs. omega ^2+omega was okay but can't believe i did not get |3-4i| till i solved and saw other procedures too!